Let \(X \subset \mathbb{R}^2\) be a connected graph that is the union of a finite number of straight line segments. Show that \(\pi_1(X)\) is free with a basis consisting of loops formed by the boundaries of the bounded complementary regions of \(X\), joined to a basepoint by suitably chosen paths in \(X\). [Assume the Jordan curve theorem for polygonal simple closed curves, which is equivalent to the case that \(X\) is homeomorphic to \(S^1\).]
Solution
1. \(X\) is a finite connected graph, so \(\pi_1(X)\) is a free group.
the fundamental group of a finite connected graph is free (a graph is homotopy equivalent to a wedge of circles).
2. The rank of \(\pi_1(X)\) equals the number of bounded complementary regions of \(X\).
2.1. Let \(V\) be the number of vertices and \(E\) the number of edges of \(X\).
setup.
2.2. \(X\) is homotopy equivalent to a wedge of \(E - V + 1\) circles.
a connected graph with \(V\) vertices and \(E\) edges is homotopy equivalent to a wedge of \(E - V + 1\) circles (collapse a maximal tree).
2.3. By Euler’s formula for planar graphs, the number of bounded complementary regions is \(E - V + 1\).
for a connected planar graph, \(V - E + F = 2\) where \(F\) is the number of faces (including the unbounded one); the number of bounded faces is \(F - 1 = E - V + 1\).
2.4. Hence the rank of \(\pi_1(X)\) equals the number of bounded complementary regions.
2 and 3.
3. The boundary of each bounded complementary region is a simple closed polygonal curve, hence a loop in \(X\).
the boundary of a bounded face of a planar graph is a simple closed curve (by the Jordan curve theorem for polygonal curves).
4. These boundary loops, joined to a basepoint by paths in \(X\), form a free basis of \(\pi_1(X)\).
4.1. The boundary loops of the bounded regions are independent in \(\pi_1(X)\).
they correspond to the generators of the free group (each bounded region contributes one generator).
4.2. They generate \(\pi_1(X)\).
the number of bounded regions equals the rank (<1>2), and the boundary loops are independent, so they form a basis.
5. Q.E.D.
4.